Consider the real matrix $$ A=\begin{pmatrix}a & b\\c & d\end{pmatrix}. $$ It is said that the real parts of the Eigenvalues of $A$ are all negative if $$ a+d<0\text{ and }ad-bc>0, $$ i.e. $$ \text{trace }A<0\text{ and }\text{det }A>0. $$

How to verify this?

If I consider the characteristic polynomial of $A$ which is $$ \lambda^2-(a+d)\lambda+ad-bc=0, $$ this has solutions $$ \lambda_{1,2}=\frac{a+d}{2}\pm\sqrt{\frac{(a+d)^2}{4}-(ad-bc)} $$

I think, we now have two possibilities.

(1) $\lambda_{1,2}$ are real if $$ ad-bc\leq\frac{(a+d)^2}{4} $$

In this case, we have $$ \lambda_{1,2}\leq a+d $$

and we need $a+d<0$ to have negative real parts.

(2) $\lambda_{1,2}$ are complex.

This is the case if the expression under the root is negative what only can be the case if $ad-bc>\frac{(a+d)^2}{4}>0$. The real part then is $\frac{a+d}{2}$ and this is negative exactly when $a+d<0$.

Summarizing both cases, we have the two conditions $$ a+d<0\text{ and }ad-bc>0. $$

  • $\begingroup$ What is your Question?? $\endgroup$ – tattwamasi amrutam Dec 30 '16 at 14:58
  • $\begingroup$ My question is if my Argumentation is correct. $\endgroup$ – John_Doe Dec 30 '16 at 15:02

First deal with the case where the two eigenvalues $x$ and $y$ are real. In this case $det(A)=xy>0$ implies that $x$ and $y$ are of the same sign, while $Tr(A)=x+y<0$ implies they are both negative.

Now if $x$ and $y$ are nonreal they are conjugated (as they are both roots of the characteristic polynomial). In particular they have the same real part, which is negative, since $Tr(A)<0$ is twice the common real part of $x$ and $y$.


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